3.2.24 \(\int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [124]

3.2.24.1 Optimal result

Integrand size = 16, antiderivative size = 46 \[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f} \]

output

arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)
 

3.2.24.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f} \]

input

Integrate[1/Sqrt[a + b*Tan[e + f*x]^2],x]
 

output

ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(Sqrt[a - b] 
*f)
 

3.2.24.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4144, 291, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a+b \tan (e+f x)^2}}dx\)

\(\Big \downarrow \) 4144

\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}}\)

input

Int[1/Sqrt[a + b*Tan[e + f*x]^2],x]
 

output

ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(Sqrt[a - b] 
*f)
 

3.2.24.3.1 Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4144

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> 
With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f)   Subst[Int[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 

3.2.24.4 Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \,b^{2} \left (a -b \right )}\)	\(67\)
default	\(\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \,b^{2} \left (a -b \right )}\)	\(67\)

input

int(1/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

output

1/f*(b^4*(a-b))^(1/2)/b^2/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*ta 
n(f*x+e)^2)^(1/2)*tan(f*x+e))
 

3.2.24.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.72 \[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {\sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a - b\right )} f}, \frac {\arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right )}{\sqrt {a - b} f}\right ] \]

input

integrate(1/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

output

[-1/2*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^ 
2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1))/((a - b)*f), a 
rctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e)))/(sqrt(a - b) 
*f)]
 

3.2.24.6 Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]

input

integrate(1/(a+b*tan(f*x+e)**2)**(1/2),x)
 

output

Integral(1/sqrt(a + b*tan(e + f*x)**2), x)
 

3.2.24.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Exception raised: ValueError} \]

input

integrate(1/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

3.2.24.8 Giac [F]

\[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]

input

integrate(1/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

output

sage0*x
 

3.2.24.9 Mupad [B] (verification not implemented)

Time = 11.85 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\sqrt {a-b}}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}\right )}{f\,\sqrt {a-b}} \]

input

int(1/(a + b*tan(e + f*x)^2)^(1/2),x)
 

output

atan((tan(e + f*x)*(a - b)^(1/2))/(a + b*tan(e + f*x)^2)^(1/2))/(f*(a - b) 
^(1/2))